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How can three resistors of resistances 2Ω, 3 Ω, and 6Ω be connected to give a total resistance of (i) 4 Ω, (ii) 1 Ω ?​

Answer» <html><body><p><strong>Answer:</strong></p><p>To obtain a <a href="https://interviewquestions.tuteehub.com/tag/total-711110" style="font-weight:bold;" target="_blank" title="Click to know more about TOTAL">TOTAL</a> <a href="https://interviewquestions.tuteehub.com/tag/resistance-1186351" style="font-weight:bold;" target="_blank" title="Click to know more about RESISTANCE">RESISTANCE</a> of 4 Ω from three resistors of given resistances, Firstly, <a href="https://interviewquestions.tuteehub.com/tag/connect-11879" style="font-weight:bold;" target="_blank" title="Click to know more about CONNECT">CONNECT</a> the <a href="https://interviewquestions.tuteehub.com/tag/two-714195" style="font-weight:bold;" target="_blank" title="Click to know more about TWO">TWO</a> resistors of 3Ω and 6Ω in parallel to get a total resistance of 2Ω which is less than the lowest individual resistance. Hence, the total resistance of the circuit is 4 Ω. ... ⇒ 1/R = 1 Ω.</p><p><strong>Explanation:</strong></p><p>hope it was <a href="https://interviewquestions.tuteehub.com/tag/help-1018089" style="font-weight:bold;" target="_blank" title="Click to know more about HELP">HELP</a> full</p></body></html>


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